I think that that's sort of the appeal of these sorts of problems.  In and of themselves, they're not particularly difficult, but they require you to think outside of the box in order to get the solution.  Once you see that light-bulb, the solution is obvious and elegant.

On your second point, yes, most of the problems could be done at school back in the day.  I don't know the state of British mathematics education, but American maths education is in a sorry state.  Most people can't wrap their head around the fact that 1/3 is in fact larger than 1/4.

I see that cos(x/2^1) = cos(2\*x/2^2) = 1 - 2sin^2(x/2^2). 
When k = 1, the product is cos(x/2^1).
When k = 2, the product is (1 - 2sin^2(x/2^2))\*cos(x/2^2).
When k = 3, the product is (1 - 4sin^2(x/2^3)cos^2(x/2^3))*(1 - sin^2(x/2^3)\*cos(x/2^3).
Does this lead to finding any useful pattern?